How do you solve x^4-7x^2+12=0?

Answer 1

x^4-7x^2+12
=x^4-4x^2-3x^2+12
=x^2(x^2-4)-3(x^2-4)
=(x^2-4)(x^2-3)
=(x+2)(x-2)(x^2-3)=0

x=-2, x=2, x= -sqrt3, x=sqrt3

Answer 2

x^4 - 7x^2 + 12 = 0

This is actually a quadratic in disguise; all you have to do is split up x^4 into x^2 and x^2. We're going to have this form:

(x^2 + ?)(x^2 + ?) = 0

Treating this like a quadratic, we know that -4 and -3 multiply to make +12, and add to make -7. So we have

(x^2 - 4)(x^2 - 3) = 0

Solving these individually, we get

x^2 - 4 = 0
x^2 - 3 = 0

(x - 2)(x + 2) = 0
x^2 - 3 = 0

x = {-2, 2}
x = {-sqrt(3),sqrt(3)}

So our four solutions are

x = {-2, 2, -sqrt(3), sqrt(3)}

Answer 3

x^4-7x^2+12=0
(x^2-3)(x^2-4)=0
(x^2-3)(x+2)(x-2)=0

Answer 4

take x^2 as (A)and then substitute the equation which now becomes
A^2-7A+12=0

NOW SOLVE IT LIKE A QUADRATIC EQUATION.

Answer 5

the easiest is to just plot it in a graphing calculator and find out where it intersects the x axis.